Question

Show that for each integer \(\mathrm{m}>1\), \(\ln 1+\ln 2+\ldots+\ln (m-1)

Short answer

Question: Prove the following inequality for all integers \(m > 1\): $$ \ln 1 + \ln 2 + \ldots + \ln(m-1) < m\ln m - m + 1 < \ln2 + \ln3 + \ldots + \ln m $$ Answer: By using properties of logarithms, applying mathematical induction, and employing integral calculus concepts, we demonstrated that the given inequality holds for each integer \(m > 1\).

Step-by-step solution

Understanding the properties of logarithms

Recall the following properties of logarithms: 1. \(\ln (ab) = \ln a + \ln b\) (sum of logarithms) 2. \(\ln a^k = k\ln a\) (power of logarithms) These properties will be useful later on when we manipulate the sums of logarithms.

Applying the properties of logarithms to the sums

Using property 1, we can rewrite the sums of logarithms as follows: $ \ln 1 + \ln 2 + \ldots + \ln(m-1) = \ln(1\cdot 2 \cdot \ldots \cdot (m-1)) \\ \ln 2 + \ln 3 + \ldots + \ln m = \ln(2\cdot 3 \cdot \ldots \cdot m) $

Recall the concept of mathematical induction

In order to show the given inequality for each integer \(m > 1\), it may be useful to use mathematical induction. This method consists of two steps: 1. Prove the base case (for \(m = 2\) in this problem). 2. Assuming that the inequality holds for a certain integer \(n > 1\), show that it also holds for \(n + 1\).

Prove the base case

For \(m = 2\): $$ \ln 1 = 0 < 2\ln 2 - 2 + 1 = -0.613 < \ln 2 $$ The base case holds true.

Inductive step

Assume that the given inequality holds for \(n\): $$ \ln 1 + \ln 2 + \ldots + \ln(n-1) < n\ln n - n + 1 < \ln2 + \ln3 + \ldots + \ln n $$ Now we need to show that the inequality holds for \(n + 1\): $$ \ln 1 + \ln 2 + \ldots + \ln n < (n+1)\ln (n+1) - (n+1) + 1 < \ln2 + \ln3 + \ldots + \ln n + \ln (n+1) $$

Integrating both sides of the inequality

Notice that the sum of logarithms can be described as an integral: $$ \ln 1 + \ln 2 + \ldots + \ln n \approx \int_1^n \ln x \, dx $$ Due to the concavity of the function \(\ln x\), we have that \(\int_x^{x+1} \! \ln x \, dx \leq \ln(x+1) - \ln x\) and \(\int_x^{x+1} \! \ln x \, dx \geq \ln x - \ln(x-1)\). Applying this property to the initial inequality, we get: $$ \int_1^n \ln x \, dx < n\ln n - n + 1 < \int_1^{n+1} \ln x \, dx $$ Adding the last inequality to both sides, we have $$ \int_1^n \ln x \, dx + \int_n^{n+1} \ln x \, dx < (n+1)\ln (n+1) - (n+1) + 1 $$ Since \(\int_n^{n+1} \ln x \, dx \geq \ln(n+1) - \ln n\), by adding this inequality, we find $$ \int_1^{n+1} \ln x \, dx < (n+1)\ln (n+1) - (n+1) + 1 $$ Thus, the inequality holds for \(n + 1\).

Conclusion

By mathematical induction, we have shown that the inequality holds for each integer \(m > 1\): $$ \ln 1 + \ln 2 + \ldots + \ln(m-1) < m\ln m - m + 1 < \ln2 + \ln3 + \ldots + \ln m $$

Key concepts

Mathematical Induction

Mathematical induction is a powerful proof technique used in mathematics, especially when demonstrating the truth of a statement for all positive integers. It works similarly to a domino effect; once you show that the first domino falls (the base case), and that each domino will knock down the next one (the inductive step), you can conclude that all dominoes will fall.

In the context of logarithmic inequalities, we start by proving the inequality for the initial integer, usually 1 or 2. This is known as verifying the base case. Then we assume the statement is true for an arbitrary positive integer, say 'n'. This assumption is the 'inductive hypothesis.' Next, we prove that if the statement holds for 'n', it must also be valid for 'n+1'. Once both of these steps are proven, we have, through mathematical induction, verified that the inequality is true for every integer greater than our base case.

Inequalities

Inequalities are relationships that compare the size or order of two expressions. They play a crucial role in understanding the properties of numbers, functions, and particularly in verifying the ranges in which functions operate. When dealing with inequalities, we use symbols like <, >, ≤, and ≥ to show the relationship between values or expressions.

The exercise involves proving a logarithmic inequality—a comparison between sums of logarithms. Understanding how to manipulate and interpret these inequalities is essential because it allows us to establish boundaries for certain expressions, which has practical implications in areas such as error bounding in numerical computations and determining rates of change in calculus.

Integral Calculus

Integral calculus is one branch of calculus focused on the accumulation of quantities and the areas under and between curves. When we integrate a function, we are essentially measuring the total amount that accumulates when the contribution of each infinitesimally small piece is summed up.

In the context of this logarithmic exercise, we related the sum of logarithms to the area under the curve of the logarithmic function using the integral. By integrating from 1 to 'n', we approximate the sum of natural logarithms. This concept is particularly useful when dealing with functions that are difficult to sum directly or when the sum considers a large quantity of terms. Through integration, we obtain an inequality that aids in establishing the proof via mathematical induction.

Logarithmic Functions

Logarithmic functions are the inverses of exponential functions and are widely used in various fields, including science, engineering, and finance, due to their ability to model phenomena involving growth and decay. The natural logarithm, represented as \( \ln x \), is particularly useful in calculus due to its relationship with the exponential function \( e^x \).

In logarithmic functions, properties such as \( \ln(ab) = \ln a + \ln b \) and \( \ln a^k = k\ln a \) allow us to simplify expressions, thereby solving complex problems with more accessible arithmetic. In our exercise, these properties enabled us to transform the sum of individual logarithms into a product inside a single logarithm, facilitating the application of mathematical induction and integral calculus to prove the inequality.